Question

is conservative 2PL is recoverable schedule ?

Comments

@Prateek Raghuvanshi i think you are using the definition of conservative 2PL i.e. it lock all the variable before it start and release  all variable after commit

but that is not so watch this https://cs.stackexchange.com/questions/51763/does-the-conservative-two-phase-locking-protocol-produce-cascadeless-schedules they are saying is is not cascadeless 

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No,

“conservative 2PL (or static 2PL) requires a transaction to lock all the items it accesses before the transaction begins execution, by predeclaring its readset and write-set”

It is not mandatory to commit before unlocking the resource. Hence, Recoverability is not guaranteed.

Please correct me if I my understanding is incorrect.

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schdeule s: r2(b),r1(a),w2(b),r1(b) is this schdeule conservative 2pl implementable?

Answer 1

No,

“conservative 2PL (or static 2PL) requires a transaction to lock all the items it accesses before the transaction begins execution, by predeclaring its readset and write-set”

It is not mandatory to commit before unlocking the resource. Hence, Recoverability is not guaranteed.

Please correct me if I my understanding is incorrect.

Comments

Yes you are correct

Answer 2

there is chance of dirty read in 2PL conservative approach so we cant say it is recoverable.recoverable schedules are generated by rigorous 2PL protocol not conservative.

reference- https://www.youtube.com/watch?v=VwZ58AvwYb0&t=532s

Answer 3

No,

In conservative 2PL, there is no growing phase so all transactions need to be aware of their future need and there is only a shrinking phase so, No need to commit before unlocking all locks. The transitions can unlock their locks if no need.

that's why there is a dirty read possible so we can’t say that the schedule is recoverable and cascadeless.